Question

\( \int \frac{dx}{12\cos x + 5\sin x} = \)

• A. \( \frac{1}{13}\log\left|\tan\left(\frac{\pi}{4} + \frac{x}{2} - \frac{1}{2}\tan^{-1}\frac{5}{12}\right)\right|+c \)
• B. \( \frac{5}{12}\log\left|\tan\left(\frac{\pi}{4} + \frac{x}{2} - \frac{1}{2}\tan^{-1}\frac{5}{12}\right)\right|+c \)
• C. \( \frac{1}{13}\log\left|\tan\left(\frac{\pi}{4} + \frac{x}{2} + \frac{1}{2}\tan^{-1}\frac{5}{12}\right)\right|+c \)
• D. \( \frac{5}{12}\log\left|\tan\left(\frac{\pi}{4} + \frac{x}{2} + \frac{1}{2}\tan^{-1}\frac{5}{12}\right)\right|+c \)

Hint:

To integrate \( \int \frac{dx}{a\cos x + b\sin x} \): 1. Write \( a\cos x + b\sin x = R\cos(x-\alpha) \) or \( R\sin(x+\alpha) \). Let \( a = R\cos\alpha, b = R\sin\alpha \). Then \( R = \sqrt{a^2+b^2} \) and \( \tan\alpha = b/a \). 2. The integral becomes \( \frac{1}{R} \int \sec(x-\alpha) dx \) or \( \frac{1}{R} \int \operatorname{cosec}(x+\alpha) dx \). 3. Use standard integrals: \( \int \sec u du = \log|\tan(\frac{u}{2}+\frac{\pi}{4})| + C \) and \( \int \operatorname{cosec} u du = \log|\tan(\frac{u}{2})| + C \).

Answer 1

The Correct Option is A

We have \( I = \int \frac{dx}{12\cos x + 5\sin x} \).
Let \( 12 = R\cos\alpha \) and \( 5 = R\sin\alpha \).
Then \( R = \sqrt{12^2+5^2} = \sqrt{144+25} = \sqrt{169} = 13 \).
And \( \cos\alpha = \frac{12}{13}, \sin\alpha = \frac{5}{13} \).
So \( \tan\alpha = \frac{5}{12} \implies \alpha = \tan^{-1}\left(\frac{5}{12}\right) \).
The denominator becomes: \( 12\cos x + 5\sin x = R\cos\alpha\cos x + R\sin\alpha\sin x = R(\cos x \cos\alpha + \sin x \sin\alpha) = R\cos(x-\alpha) = 13\cos(x-\alpha) \).
So, \( I = \int \frac{dx}{13\cos(x-\alpha)} = \frac{1}{13} \int \sec(x-\alpha) dx \).
We know \( \int \sec u du = \log|\sec u + \tan u| + C \).
Or \( \int \sec u du = \log\left|\tan\left(\frac{u}{2} + \frac{\pi}{4}\right)\right| + C \).
Let \( u = x-\alpha \).
\[ I = \frac{1}{13} \log\left|\tan\left(\frac{x-\alpha}{2} + \frac{\pi}{4}\right)\right| + c \] \[ I = \frac{1}{13} \log\left|\tan\left(\frac{x}{2} - \frac{\alpha}{2} + \frac{\pi}{4}\right)\right| + c \] Substitute \( \alpha = \tan^{-1}\left(\frac{5}{12}\right) \): \[ I = \frac{1}{13} \log\left|\tan\left(\frac{\pi}{4} + \frac{x}{2} - \frac{1}{2}\tan^{-1}\frac{5}{12}\right)\right| + c \] This matches option (1).